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Solutions for Session 8, Part B
See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6| B7 | B8 | B9 | B10 | B11 | B12 | B13
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Problem B1 | |
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Here is the completed table:
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Number of teachers | | Amount each will receive |
1 | | 800,000 |
2 | | 400,000 |
3 | | 266,666.66 |
4 | | 200,000 |
10 | | 80,000 |
15 | | 53,333.33 |
20 | | 40,000 |
100 | | 8,000 |
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The last two entries will vary.
<< back to Problem B1
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Problem B2 | |
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One rule is that the product of the input and the output is always 800,000. If M = the money each teacher receives, and T = the number of teachers, then (M) x (T) = 800,000. This is because the total prize of 800,000 is split evenly among the teachers. This also leads to a second rule: M = 800,000 / T, because the money received by each teacher is 800,000 divided by how many teachers split the prize. Additionally, T = 800,000 / M.
<< back to Problem B2
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Problem B3 | |
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The amount of money decreases as the number of teachers increases, but it is not an exponential decay because the ratio between consecutive outputs is not constant. Looking at differences between outputs guarantees that the graph is neither linear nor quadratic. The graph is not cyclic, because there is no point where the outputs begin to repeat.

<< back to Problem B3
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Problem B7 | |
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Overall, if x is multiplied by a number, y is divided by the same number. If x is divided by a number, y is multiplied by the same number.
a. | Here is the completed table. All y-values are rounded to one decimal place. |
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x | | y | | Decrease in y |
20 | | 100 | | -- |
30 | | 66.7 | | 33.3 |
40 | | 50 | | 16.7 |
50 | | 40 | | 10 |
60 | | 33.3 | | 6.7 |
70 | | 28.6 | | 4.7 |
80 | | 25 | | 3.6 |
90 | | 22.2 | | 2.8 |
100 | | 20 | | 2.2 |
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b. | As x increases by 10, y continuously decreases, but the rate of decrease lessens as x grows. |
c. | If x doubles, y is cut in half. If x triples, y is divided by three. |
d. | If x is very small, y must be very large, since the product of x and y is always the same. By the same argument, if x is very large, then y must be very small. |
<< back to Problem B7
<< back to Problem B7 low-tech version
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Problem B8 | |
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Let's try it:
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x | | y |
-10 | | -0.3 |
-6 | | -0.5 |
-3 | | -1 |
-2 | | -1.5 |
-0.5 | | -6 |
0.5 | | 6 |
2 | | 1.5 |
3 | | 1 |
6 | | 0.5 |
10 | | 0.3 |
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<< back to Problem B8
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Problem B9 | |
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If x = 0, y is undefined, which means that no value of y will make zero times y equal to 3. Using a calculator, 3 / 0 will fail to return a number; the calculator will give an error message.
<< back to Problem B9
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Problem B10 | |
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The graph will not cross the y-axis, because if it did, it would mean that some y-value would be assigned for x = 0 for the graph, and, as explained in the solution to Problem B9, there is no such value. Another way to look at it is to examine the behavior of the graph near x = 0. If x is a little more than zero, y is a very large, positive number, but if x is a little less than zero, y is a very large, negative number. These portions of the graph do not meet.
<< back to Problem B10
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Problem B11 | |
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The 3 represents the area of a rectangle drawn with (0, 0) as one corner and any point on the graph as the opposite corner. Compare this to the Interactive Activity, where there were rectangles of area equaling 2,000 feet -- length times width always equaled 2,000. Here, x is the length, and y is the width.
<< back to Problem B11
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Problem B12 | |
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In a direct variation function, an increase in x creates a proportional increase in y; if x is multiplied by 5, y is also multiplied by 5. But in inverse variation, the opposite is true: If x is multiplied by 5, y is divided by 5. The word "inverse" refers to the inverse operations of multiplication and division.
<< back to Problem B12
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Problem B13 | |
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In this particular equation, x is the reciprocal of y, because x and y multiply together to make 1. The reciprocal is used to solve equations like 5n = 16 -- multiplying both sides of the equation by the reciprocal of 5 produces two numbers (5 and 1/5) which, when multiplied, make 1. So, multiplying by 1/5 will remove the 5 from the left side, leaving variable n by itself. This is the key to solving equations by backtracking.
<< back to Problem B13
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